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What bandwidth and availability is required for modern air traffic control?
A certain telecoms provider is advertising in the UK by showing off that you can operate air traffic control over it's home broadband service.
Is this that surprising?
I didn't think that air traffic control for commercial flights was that bandwidth hungry or relied on an ultra stable connection?
What bandwidth (in kb, mb, gb etc. per second), and how many nines of availability are required?

I think 'ultra stable' is required, you want air traffic control to be immediately available when needed. When there is an issue onboard, the airplane has to land immediately, and the runway needs to be cleared, traffic control cannot be offline for a few minutes.

The entire Chinese railway system (one of the largest, fastest and most busy) run on GSM for two decades before switching to 4G and NB-IoT. Here I'd trust wireless much more than any landline. While each node and switch along the line could fail, the cable itself isn't 100% reliable either. Wireless has much fewer points of failure.

There 31 million seconds in a year.  Five-9s (99.999% the industry typical value) reliability would be about 5 minutes of downtime per year.  A 5 minute outage at the exactly wrong-moment could be catastrophic. At any randomly chosen time, even a 1 minute outage could easily lead to separation loss.  I would estimate that Five-9s is not enough reliability for ATC service.

I think you’ll have to be more specific. There are a lot of different ATC roles that may be more or less suited to being deployed remotely.  I could easily see clearance delivery being done that way.

Remote towers are already in use at several sites in multiple countries. Off the top of my head and without even googling, there's Leesburg airport in Virginia, US (JYO), an airport in Colorado, and one of the London-area airports I think. Bandwidth for towers (which require video) will be massively greater than bandwidth for radar position data. As everyone has said, availability (uptime) is the relevant metric. See also: virtual ATC networks (PilotEdge, VATSIM, IVAO) which are doing this already, though with much fewer aircraft and absolutely zero real-world consequences for failure.

@user3528438 Wireless has many more points of failure. But that isn't important. What is important are Single Points of Failures. Wireless systems can have redundancy added very easily.

@abelenky, Per The Canadian Automated Air Traffic System (CAATS): An Overview, availability requirements are "0.9997 for full service mode to 0.999997 (1 sec/wk 95sec/yr) for an emergency mode of operation". Depends on how you measure those functions. Paper lists a multitude of services within ATS. #1 is Comms, (radio/phone) not Track, and is an entirely separate system to radar / tracking flight data systems.

You can, technically, but you wouldn't run ATC over a home broadband service. The problem is not the bandwidth of the broadband, but the availability (is it available when you want to use it) and continuity (the probability that keeps being available while you are using it) are too low, and the Mean Time To Repair is too high.
In terms of bandwidth for radar control, you don't need much (20 - 200 kbps for a radar, 500 kbps for radar tracker output with 250 aircraft and 1 second update rate) nor for Voice over IP (100 kbps per active channel). But if you operate a remote tower, such as in this case, you need to be able to stream a few 4K video feeds with low latency (<1 second end-to-end)(~20Mbps), which may be a bit of a challenge for some networks. In addition, in case of degradation of the bandwidth, you want to be able to prioritise voice over video for example.
Here is the making off:

Anyway, no air traffic control was done in this video. Notice the ground station call sign "air suburbia ops" and the phrase "runway 23 available for landing". If anything was actually transmitted from that living room (we don't know, we are looking at cut/edited scenes), it was purely informational, not air traffic control.

Actually very little bandwidth is required for most Aviation control related items. This is due to the substantial redundancy built into the "entire system."  Each part plays a role in ensuring adequate control of the air traffic in the associated area.  The FAA has over 30,000 stations around the US.  They range from VOR's to weather camera's to AWOS (Automated Weather Observation Stations) sites, to regional ZAN's/ATC's.  Regional ATC's require the most bandwidth but the FAA is just now getting out of the TDM world and moving to ethernet.
TDM, aka legacy circuits such as T1/DS1's, DS3's, and OC-3's, provided extremely predictable bandwidth that was completely serialized.  The FAA took advantage of the serial nature of TDM and built an entire network of equipment, in excess of 150,000 unique service/location combinations, and it still works to this day.  When digital bandwidth was expensive in the 1970's the FAA built most of its services to rely on 9600 bits/sec.  Some things rely on only 2400 bps.  With a 9600 bps requirement they could stack 5-6 services on a single 64k digital channel.  This saved the FAA alot of money over the years.
What is more interesting is the detail the FAA has on each site.  Of the 30,000 plus locations the FAA has on record they have a strict listing of the redundancy requirements, which generally translates into uptime requirements, for each one.  In some locations a single circuit is ok.  Other sites two circuits/connections are ok.  In other sites two connections are required but they must be from different providers.  In other sites the connections must be different providers, physically diverse from each other and terminate at different Central Offices. Going even further the FAA might require physically different transports or even three connections, all of which are physically diverse in all aspects.  In the legacy TDM days the FAA would request a circuit path that would require the circuit to avoid certain central offices even when going across a small town.  In some regional centers there will be multiple carriers with multiple physically diverse entrances for the remote connections.
That said the question is related to Traffic Control. The FAA needs very little bandwidth for traffic control. There are larger pipes, OC3's and OC12's, now moving to Gigabit ethernet, between the regional centers so they can share radar and traffic control data but generally speaking regional centers can operate independently, though that is far from ideal. Most of the data is not air traffic control though and even the centers migrating to gigabit ethernet are doing it for the cost savings, not the bandwidth needs. Telco's are raising their prices substantially for TDM (T1, OCx) services making it prohibitively expensive to stay on TDM Circuits.  For voice traffic to the airplanes this is minimal.  G.711 is ~90k when active but there are codecs that are more efficient. Most airline carriers have large networks that consist of remote SIP-to-RF radio's but most are single channel radio's so they would only need less than 100k when they are active.
That said the system has a massive amount of redundancy built into it to ensure that no one location knocks out the entire system.  This is matched with the rules of flying that generally state the pilot is the one in control of the airplane and is responsible to be situationally aware at all times.
I know all of this because I work for a Telecom in Alaska that supports the FAA. Alaska has been a proving ground for the next generation equipment remote equipment for the FAA.  Alaska's vast and remote distances pose significant challenges to the FAA in terms of reliability, both the equipment itself as well as the communications to that equipment, and the criticality of the equipment. In fact ADS-B was designed and tested in Alaska extensively before it was rolled out nationwide.
All said and done the bandwidth is not the issue for the FAA.  Generally speaking control data is very little, be it in Aviation or controlling an oil field.  The sheer number of sensors can add up, but it is nothing compared to the data a Netflix PoP pushes in a day.

Really nice answer! Welcome to Av.SE, Scotth.

I'll give it a shot...  I don't think a bandwidth value can be provided because it's a distributed and decentralized system, so it's bandwidth on many channels, not bandwidth on a single fibre-optic cable running to someone's house.
So it's a nonsensical ad from the bandwidth perspective.  ATC is mostly radars, with computers processing the radar information and transmitted data received from the airplanes (now moving to on-board GPS location transmitted to ATC by satellite for processing), for presentation and further processing by the supercomputer inside the head of the air traffic controller.
The heavy lifting in ATC is done by the unique characteristics of the brains of the people working the radar displays.  The recruiting, testing and training systems of the ATC agencies is designed to sift out these brains from the general population.  That's why previous education and experience doesn't count that much toward your chances of success when you try out for ATC.  If you have the mental skills they're looking for, you'll make it in as long as you're free of other show-stopper defects in personality or character.
Beyond a moderate amount of bandwidth in the communication networks of the ATC computers (system development for those sorts of hardened institutional uses follows many years behind development for commercial and consumer uses, and if you went to a ATC facility the hardware is likely 10-20 years old), the big priority is hardware reliability.
You don't want the human computers monitoring the hardware computers, and making all the decisions, to lose the feed for any length of time, and to that end you will achieve reliability the same why the airplanes themselves do, with redundant network and processing hardware architecture to eliminate single-points-of-failure.
So someone in the industry would see that Ad and conclude it's claiming it has a hardened and redundant network with bulletproof reliability, not so much super-duper bandwidth or speed.

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What is this plane with its thick cockpit?
Today I saw this unusual plane while going to work. I'm in France, near Paris. It flew at a rather low altitude, even if I can't tell precisely - I noticed the low altitude because of the loud sound.
We have an airport at 20km, the Charles de Gaulle Airport (CDG).
The plane was going towards South-West, probably directly departing from it. It came from this direction.
I was in the place named "La Défense", a district in the west of Paris.
I did make an outline sketch but it's not very precise :
It looked like a very small airliner; what sparked my attention was the front of the plane (the cockpit, very probably), which was thicker than the fuselage, making it look like some sort of hammerhead shark.
The wings were posistioned more towards the back of the fuselage than usual, but I can't remember if they were straight or oriented like usual.
As well it had a T-shaped tail.
I'm not very knowledgeable about planes but I have a some interest and I have never seen one like this.
I'm curious because it really looked unusual to me and that cockpit was funny.

Could it be just a canard instead of a large cockpit ?

I have tried to confirm my answer about the Avanti using flight tracking websites, but have not had luck so far. Could you narrow down the time window, geographic area or direction of flight?

I didn't know about these "canards". Indeed this could probably it, It was difficult to tell.
I'll add details about the flight

I'm surprised no-one has mentioned FlightAware. I saw a strange plane over my house, noted the time, and later looked up arrivals around that time at the nearby airport. My strange plane happened to be the same type of plane in the accepted answer.

@TKK nice tool, thank you for the recommandation

Potentially a Piaggio Avanti, based on OPs description of an unusual nose.

this really looks like it ! I'll wait a little but I thing you got it

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What is this "Supersonic French" aircraft?
To celebrate Bastille Day, Ubisoft released the below (2D) in-game charm for their game, Tom Clancy's Rainbow Six Siege.

Despite frantic searching on Wikipedia, I cannot for the life of me figure out what aircraft the charm is based on, or if indeed a real-world equivalent exists at all. At first it looked like a Rafale, but the vertical stabilizer does not seem to fit any that I could find an image of.
It's not a question of life and death, but I would very much be interested in what aircraft is actually portrayed here.

It may not be a real one at all.

the alpha jet is not supersonic.

The rafale would have canards anyway. Anything mirage would be supersonic but delta winged. They don't look like the Jaguar or the endenard/super endenard

@ManuH The naming used by Ubisoft is misleading; Alpha Jet is not indeed supersonic, but whoever responsible for naming at Ubisoft decided to name it as such.

The "Patrouille de France" (aerobatic section of French Air Force) flies over Champs-Elysées on each July 14th (aka "Bastille day" elsewhere), the national day, to close the aerial military parade. They have Alpha-Jet currently, recognizable to their pronounced anhedral wing for reduced stability.

@mins I though anhedral wing where meant to improve stability as discussed in this answer

@ManuH: "The combination of aft sweep and high-wing configuration together achieve so much stability that it has to be counteracted with an anhedral" (source, p 7-37). Many fighters have a negative dihedral angle for agility. Sweep (and twist) also affects the stability (according to Raymer 1° of dihedral is equivalent to 10° of sweep). (you can visualize the recovery effect of the dihedral/anhedral here).

I believe this is the Alpha Jet

Source of pictures: Wikipedia
It is famously used by the Patrouille de France (the French military acrobatic demonstration unit). They are most famous for overflying the Champs-Élysées during the July 14 celebration, drawing a French flag in the sky.

It is the Alphajet, and contrary to the OP, it is not supersonic

This is the correct answer

Actually contrary to Ubisoft, they named it such for whatever reason ;) Seems like it would be the Alpha Jet indeed, the vertical stabilizer is just modelled different for whatever copyright reason. Thanks!

Added some context to explain the choice. Feel free to roll back/edit.

@Antzi I left the bulk of it in there just for esoteric reasons, but it doesn't explain why I chose that aircraft. I chose it based on the appearance and that it's an aircraft used by the French military.

Actually....  pretty sure someone just traced an image of a model of the alpha jet from an online gift store for the patrouille. Sneaky sneaky :D

Re above-- nice find; deserves to be incorporated into an answer.

@JourneymanGeek Ubisoft is so lazy

I suppose this is why aerospace engineers stick to engineering and video game creators stick to creating videogames.


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Where is Ronda's airstrip, or why does Ronda have an airport code (RRA) if it has none?
I'm puzzled by Ronda, Spain having an airport code.  Ronda has no airport.  I've never seen an airstrip anywhere near either,
Wikipedia notes that some railway stations have IATA codes, but notes that The IATA codes for railway stations normally begin with Q, X or Z, except when the station shares the code with an airport.  So according to the IATA codes it does not refer to the railway station.  ITA Matrix does not know RRA.
The tourist website andalucia.com includes Ronda in a list of unsurfaced airstrips, but uniquely without coordinates.  Other websites place the airstrip in or near the city centre of Ronda, which is untrue, as a topographic map confirms.
Does Ronda have an airstrip?  If yes, where?  If no, why does it have an IATA code?

@Muzer Nice find.  How did you search?  Confimed by topographic map, Pista de aterrizaje de CMA.  So it does exist, then.  Pista de aterrizaje is runway.  But what is CMA?  Still surprising such a tiny airstrip has an IATA code.  Apparently, it is Cuerda media aerodinámica, which means Mean aerodynamic chord, which is an odd name for an airfield.

I found a photo.  Beautiful airstrip? https://goo.gl/maps/FXnKdh468y32

@gerrit: Possibly Consejería de Medio Ambiente, the Andalusian regional government's Ministry of the Environment.

@JonathanReez: [Aviation.SE] deals more in ICAO matters than in IATA ones.

IATA does not seem to agree that this code exists. It may have been decommissioned.

The short answer seems to be that Ronda has no assigned IATA code, and RRA isn't in use anyway. That's according to IATA's own directory; third-party sites aren't always reliable.
An airport was in fact proposed for Ronda in the 1980s and never built; the City rejected the license application in 1983. I found two old articles about it from El País: in July 1983 they reported a dispute between the airport's promoter, Prince Alfonso of Hohenlohe-Langenburg, and the Ayuntamiento de Ronda (the city council):

Su promotor, el príncipe Alfonso de Hohenlohe, pretendía sacarlo
  adelante con una licencia de obra menor por valor de medio millón de
  pesetas, y asegura que para montar el Marbella Club y lanzar el
  turismo en la Costa del Sol no le pidieron tantos requisitos. El
  alcalde, el socialista Julián de Zulueta, recela de intereses ocultos
  tras el proyecto, descarta la tesis de que se trate de un simple
  aeródromo deportivo y no está dispuesto a permitir su construcción si
  no es con todos los requisitos.

Its promoter, Prince Alfonso of Hohenlohe, claimed he could proceed with a construction license for less than half a million pesetas, and said that building the Marbella Club and promoting tourism in the Costa del Sol hadn't been so demanding. The Socialist mayor, Julián de Zulueta, is suspicious that there may a hidden agenda behind the project, rejects the claim that it will be a small, recreational aerodrome, and is unwilling to allow construction unless all formalities are complied with.
By October the same year, the council had formally rejected the plans:

El Ayuntamiento de Ronda decidió, en pleno extraordinario, denegar
  cualquier licencia para la polémica construcción de un aeródromo en
  las afueras de la ciudad.

In an extraordinary session, Ronda City Council decided to deny a license for the controversial construction of an aerodrome on the outskirts of the town.
The articles say the runway was intended to be 2,200m by 80m (7,200ft by 262ft) which, if true, certainly seems to be excessive for the light, recreational traffic and gliders that the Prince claimed would use it.
I couldn't find any more recent references to an airport in Ronda.

Apparently, there is an airstrip a roughly 4 km east of the town, as confirmed by an IGN map, IGN aerial photography, and Openstreetmap (Thanks Muzer).  It does not give the impression of active use; a Panoramio photo shows a field of sunflowers.  Its full name on the topographic may is Pista de aterrizaje de CMA, which means Runway of the CMA, where CMA may be Consejería de Medio Ambiente, ar the Andalucian government's Ministry of the Environment (thanks Henning Mankolm).

I seriously doubt though that that airport would have IATA code, especially since it does not even seem to have an ICAO one—ourairports identifies it as just ES-0071

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Does the APU consume a lot of fuel compared to a jet engine?
I would like to know whether the APU of modern commercial airliners consumes more or less fuel than a jet engine. Of course, this will vary between aircraft types, so what comparisons in fuel usage are there on a modern jet such as a B737 or A320? What is the ratio between an engine hourly fuel consumption and the APU hourly consumption?

Nit-picky perhaps, but keep in mind that a typical APU IS a jet engine.  It's just a smaller one.  :-)

@Lnafziger yet designed to produce only electrical and hydraulic energy, not kinetic energy as jet engines mounted underwing

Absolutely!  I would suggest rewording slightly so that it is more clear that you aren't saying that an APU is NOT a jet engine though.

To give a comparison with an engine at idle:
On the 737 NG, the APU can use 240 lb/hr when under load. It is rated at 90 kVA, same as the generator on each engine.
The engines can use about 850 lb/hr at idle, which from those figures is about 1/10 of fuel flow at full (takeoff) power. So the engines use about 4 times as much fuel (each) as the APU does to just provide power. Of course these values will vary based on ambient conditions and the load placed on the systems (like the bleed air system).
This is why if a plane has to stop away from the gate for a while, they will shut down the engines and run on the APU.

And remember that is idle engine power for an entire hour. In reality the engines will most likely be operating at some thrust and the fuel consumption is much larger.

With that fuel burn and output power. That translates to roughly 7% efficiency for the APU.

The APU consumes less fuel because it doesn't have to contribute thrust to the airframe.
Many APUs have the same generator capacity as a propulsion engine.  But they don't produce thrust. They can run at an optimum RPM for the generator requirements.
A320
                   APU
Ground operations: 130 kg/hour           90 kVA
FL200:             65
FL300:             51

On a short flight each main engine might consume an average of 1260 kg/hour.

A better comparison would be with the main engine idling and/or with sufficient RPM to produce 90 kVA.  How much does the main engine consume with that as a load?

@SkipMiller That's true, depending on what you conisder. For normal on ground APU usage, you are right. To generate power in case of in-flight loss of engines (e.g. US Airways Flight 1549), one can always quibble.

@Skip: Interesting Q. I think it would be surprising if an engine primarily designed to produce thrust would generate electricity more efficiently than an engine specifically designed for that purpose. But the APU may be optimized instead to reduce weight I guess.

even on the CFM56 commercial presentation  I cannot find consumption information

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Can private flights between the US and Canada avoid using a port of entry?
I'm considering getting back into flying private after a 15 year absence. I'm a dual US/Canadian passport holder. I'd like to fly between homes on Nantucket, MA and rural Nova Scotia regularly. Is there any type of permit that would provide an exemption from landing at a port of entry every time?

CANPASS gives a streamlined entry into Canada, allowing you to land at a port of entry and just report by phone.  To go into the US, you need an eAPIS account.  I think that with eAPIS you still need to meet in person with a customs person.  A friend in Montreal used eAPIS to travel to Vermont regularly.  He said it was a lot of hassle to get set up and go through the process the first time, but after that it was not too bad.  On the way back he just had to land at a convenient POE and call in.
Not quite an avoidance of POE requirements, but otherwise pretty convenient.

Unfortunately, no. There are simply too many people who would abuse such a permit to commit crimes for either country to accept the risk.
You will need a stop at a Port of Entry or User Fee (US) airport in each direction to clear customs and immigration, but at least it should be quick since you are a citizen of both countries, and there appear to be several along your likely route of travel to choose from. Then you can continue on as a domestic flight to wherever you want, including a private strip.
You’d think that Nantucket (ACK/KACK) would be large enough and busy enough to be at least a US Customs User Fee airport, which would solve half your problem, but apparently not.

Thank you, Stephen. I was hoping there was some qualifying-type blanket clearance that I was unaware of, but alas I understand the reasoning.

@DeanG - Don’t forget that you may need a DTOPS Decal for the aircraft as well. You will still need to land at an AOE.


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What determines the maximum altitude a plane can reach?
What factors determine the maximum altitude for a plane?
Is it limited by wing design, engine thrust, and so on?
Is there a formula by which one can calculate the maximum altitude a plane can reach?

This really depends on what type of plane. Most require oxygen to make a combustion. Future Space Planes will have a dual engine system. Propeller are limited by other factors...

weigth: aircrafts travel higher after burning some fuel, wing design: an TR1 or a RQ4 fly higher than a passenger jet, engine trust, ...

Um... could yo precise the question? Are you looking for the maximum "physical" altitude an airplane can reach (absolute ceiling as described in the answers) or are you refering to the maximum "allowed" altitude an airplane can reach (service ceiling, limited in this case by a given rate of climbing). Actually only the last answer addressed this point :)

Interplanetary Cessna https://what-if.xkcd.com/30/

The higher you get, the lower the density of the air becomes. This lower density results in a lower lift being generated for the same airspeed and angle of attack. Effectively, the higher you fly the higher your minimum speed becomes. So while climbing, your speed needs to increase to compensate for the lower air density. As long a you can fly faster, the lower density at altitude can be compensated for.
Basically there are two things that limit your maximum speed: thrust and speed of sound and with that your maximum altitude.
First is thrust; the higher you get, the lower the thrust your engines deliver. You might note that drag goes down with the air density as well but since you are flying faster and faster during the climb the drag doesn't decrease at all. If your maximum altitude is limited by thrust then at some point during the climb the thrust and drag are getting close to equal and that is where the climb stops. When you can no longer climb with more than 100ft per minute (for propeller aircraft) or 500ft per minute (for jet / turbofan aircraft) you have reached your service ceiling. If the aircraft maximum altitude is determined by thrust, the absolute ceiling will take very long to reach.
At high altitudes air breathing engines will get difficulties eventually. Due to the lower air density the mass flow through the engine is reduced up to a point where it causes a flame out.
The other limitation is the speed of sound, at least for subsonic aircraft. In the process of generating lift, air flowing over the top of the wing is accelerated. At one point, when the aircraft is still flying below the speed of sound, shock waves will start to form over the wing. This results in increase of drag and reduces the lift. So provided you have enough engine power at your disposal you can climb to an altitude where your minimum speed is also your maximum speed. This is called the coffin corner. In the coffin corner:

if you fly any faster, you will exceed the maximum Mach number ($M_{mo}$) of your aircraft, resulting in high speed buffet, vibrations and possible loss of control.
if you fly any slower, the maximum lift that the wing can provide will be insufficient to maintain altitude. Descent or the aircraft will stall.
if you fly any higher and you will be too fast and too slow at the same time.
if you turn, you increase the wing loading, thereby increasing the minimum speed needed to create the required lift. Also the outer wing will easily exceed the maximum speed while at the same time the inner wing is below stall speed. This can quickly develop into a spin.

Since accurate knowledge of engine performance, drag and wing characteristics of the aircraft is needed, there is not a simple formula to derive the maximum altitude for an aircraft.

Apart from the aircraft performance related limitations above there is a certified maximum operating altitude for the pressurized cabin. This takes into account the structural properties of the hull (pressure difference between inside and outside) and the achievable emergency descent rate in case of a depressurization event.

heh, nice description of the coffin corner: “if you fly any higher and you will be too fast and too slow at the same time.”

Rather than saying "If you fly slower, you will stall" might it be more accurate to say "If you fly slower, the steepest angle of attack that would avoid a stall will be insufficient to maintain altitude"?  A plane with upward vertical momentum one might pass the maximum altitude where it could sustain level flight, but if the pilot kept the airspeed and angle of attack within safe limits the plane would simply descend until it was in thick enough air.  If there are no mountains in the way I wouldn't think the plane should "notice" the changing vertical velocity.

@supercat I think that is more accurate indeed. The plane will "notice" changing vertical velocity though. The change is caused by an imbalance between lift and weight; the wing loading is different.

@DeltaLima: Strain gauges on the wings could detect the quantitative change in lift, but if one didn't have a calibrated accelerometer or know how much the plane weighed, and couldn't sense the ground, I don't think the behavior of the plane would qualitatively change at maximum altitude (unlike, e.g. if the plane exceeds maximum airspeed or maximum angle of attack, either of which would cause its behavior to change markedly).

@supercat If you are flying below the minimum speed for sustained level flight and keep the angle of attack at the maximum, then the weight exceeds the lift thus you are accelerating downwards. That will increase the angle of attack beyond the maximum unless you decrease the pitch angle constantly to match the change in flightpath angle. That is a qualitative change in behaviour isn't it? It is straight and level flight vs parabolic flight.

@DeltaLima: Yeah, you're right.  I forgot to consider the required change in attitude necessary to maintain stable flight.  Even so, if a pilot was trying to ride along the altitude ceiling, overshooting it slightly would merely require a minor change in attitude.  By contrast, exceeding the critical angle of attack would cause a sudden radical change of behavior which could only be reversed by a significant change in the other direction; while maximum airspeed isn't a single critical value (the more stress one is willing to accept, the faster one could fly, to the point that the plane...

...which would be able to survive for millions of hours under low-stress conditions might only survive for a few minutes or seconds before it falls apart).  In any case, I would think the primary issue with the altitude ceiling is that it doesn't represent a threshold where something bad will happen if it's exceeded, so much as it represents a limit of how high a plane will go if other parameters are kept in range.

@supercat Before you are going to try out your theory and kill yourself may I suggest you investigate high altitude operations, especially: mach stall, backside of the power curve, longitudinal stability, effect of temperature variations on mach number and effects of turbulence on speed and load factor to start with? There are very good reasons for the certified ceiling of an aircraft to be well below the absolute ceiling. The name coffin corner does not come out of the blue (well maybe it does, but I guess you get the point)

@DeltaLima: Fair point, from the perspective that safe operation of an aircraft requires that one always have an "escape plan" in case of things like unexpected turbulence, and simultaneously pushing the limits on two parameters at once could leave one in trouble if the air conditions into which one flies aren't exactly as expected.  Still, I would think (correct me if I'm wrong) that if one avoids pushing the limits for airspeed or angle of attack, the fact that the airframe reached an altitude where it would no longer accelerate upward would not make anything particularly dramatic happen.

Let us continue this discussion in chat.

The maximum altitude is limited by a number of factors, and the one which counts depends on the particular aircraft. These are:

Engine power output. Airbreathing engines produce less power the higher they operate due to decreasing density with altitude. In reciprocating engines, this can be overcome with turbocharging, and dedicated high-altitude piston engines use triple-stage turbochargers with intercoolers. In dedicated high-altitude designs, the engine is the smallest part of the propulsion package, most is cooling and ducting. The propeller has to be matched to the low density at high altitude, increasing in diameter for operation in low density air.
Combustion chamber pressure: The altitude limit of jet engines is mostly determined by the pressure ratio of the intake and the compressor. If this pressure drops below the minimum for sustained combustion, the engine will flame out. Since jet engines are in principle a big turbocharger where the piston engine has been replaced by a combustion chamber, this combustion chamber becomes the weak link.
Wing loading: The lower the wing loading, the lower air density can become before a wing will fail to produce enough lift. If the engines produce enough power for sustained flight (electric propulsion with solar panels, for example), the limit becomes the structural integrity of the light structure. See this answer for an applied example.
Maximum flight Mach number: For supersonic aircraft, the limit is given by a combination of wing loading and maximum speed. The faster the aircraft can fly, the lower air density can be. In most cases, the speed limit is given by intake efficiency, because intakes need to be optimized for their flight Mach number, and thermal limits due to airframe heating. Note that a fast aircraft with lift reserves can perform a pull-up at altitude, converting kinetic energy into potential energy (aka altitude), so the instationary maximum altitude could be several 1000 m above the stationary altitude limit.
Aerodynamic efficiency: This is the only factor where I can give you a simple equation, and it is determined by the aerodynamic quality of the wing and its airfoil. It applies to subsonic flight where an increase above a critical flight Mach number will reduce lift. Expressed as the minimum air density $\rho_{min}$, this is $$\rho_{min} = \frac{2\cdot m\cdot g}{(Mach^2 \cdot c_L)_{max}\cdot a^2\cdot S}$$

Here we find again wing loading $\frac{m}{S}$ as a factor, but also the maximum of the product of the square of the flight Mach number $Ma^2$ and the lift coefficient $c_L$. $a$ is the speed of sound. A good value of $Mach^2 \cdot c_L$ is 0.4, and it needs supercritical airfoils to be achieved. Use this number for modern designs and you will get a pretty accurate answer if engine thrust is sufficient. For older designs, values between 0.3 and 0.35 are a better fit. Very early designs with poor aerodynamics like the Westland Welkin would only achieve a $Mach^2 \cdot c_L$ of below 0.2.

In the most basic form, the aircraft's max altitude is the point where thrust required is equal to thrust available. This compares the thrust required to maintain airspeed and altitude to the thrust available from the engines. Since air-breathing engines will tend to produce less thrust as altitude increases, this means that the thrust available decreases with altitude. At some point, the aircraft will be at the lowest drag possible in level flight, and using all thrust available.
As casey points out, there will be many other factors, such as the ability of the aircraft to remain pressurized, the ability of the engine to maintain a certain thrust level, and atmospheric conditions.
However, if you are looking for the absolute max altitude it's able to reach, but not necessarily maintain, it becomes much more complicated. This would be determined by the maximum amount of energy an aircraft can attain, both in altitude and airspeed. An aircraft may be able to dive or remain at a lower altitude to gain speed, and then climb to trade that speed for altitude, reaching a higher altitude than it can maintain (see this incident for an example of an aircraft that flew to a higher altitude than it could actually maintain).

Did Flight 3701 actually exceed FL410 by any significant amount? If I read the Wikipedia account correctly, the flight leveled off at FL410, but the excessive climb rate caused its airspeed to be low. So the narrative, "traded speed for altitude and then could not maintain that altitude" is literally true, but  I didn't see any indication in the article that the aircraft actually climbed higher than the maximum altitude it could have maintained if the pilots had set a lower climb rate to get there. But maybe there's something that I missed or that Wikipedia left out.

It's true that it technically could fly at FL410, but the point is more that they reached that altitude in a state that was not sustainable.

The absolute maximum altitude a plane can reach is only limited by the lift it can produce.   This will be a function of the wing (and one of our resident engineers can explain this) and the airflow over the wing.  The airflow in turn is a function of your altitude (air density) and airspeed.  Airspeed in turn is a function of your thrust, drag, etc.  In short, the lift you can produce is dependent on a lot of things indirectly and this will define the physical limit of max altitude.
Note that the max altitude defined by your lift is a max continuous altitude.  If you have the momentum available you could use that to climb above this altitude for brief excursions, but you would not be able to maintain altitudes above this limit.
Note that this altitude is not the airplanes service ceiling, which is going to be lower due to climb rate thresholds (e.g. 100 fpm) or certification issues (e.g. 25,000 feet for pressurization/oxygen requirements).

Limited by both lift and thrust, not "only lift" (although you could argue that there is no lift without thrust). Your engines need to be able to operate in the low density, low oxygen environment as well...

An engineless aircraft is not subject to two of the five limiting factors in Peter Kämpf's comprehensive answer.  The current subsonic level-flight altitude record is held by the Perlan II glider which reached 76,124 feet in September 2018, exceeding the U2's record of 73,737 feet.  If Perlan II reaches its design altitude limit of 90,000 feet it will exceed the SR-71's (supersonic) level-flight altitude record of 85,068 feet.
Perlan II, while highly specialised and having a pressurised fuselage, is not dramatically different in appearance from an open-class glider.  The most significant difference is in the airfoil, which is optimised for flight at 60,000 feet.  This also results in a significantly wider range of airspeed at extreme altitudes (the 'coffin corner' referred to in other answers) than the U-2, which at operational altitude had only a 5-knot flyable airspeed range.


text:
Why is the centre of pressure for a real wing with low aspect ratio more forward than that in two-dimensional flow?
Why is the centre of pressure of a finite wing ahead of the quarter chord point predicted by 2D flow over the airfoil? Also what is a "LOW" aspect ratio in terms of numbers. The literature has terms like "very low A.R" "low A.R" "Moderate A.R" and "High A.R", but the actual ratios in terms of numbers are not given.

very low aspect ratio denotes a slender body where its length l is far bigger than its width b: l >> b. Think fuselages, external tanks, projectiles and the like. For wings of varying chord over span, it is preferred to use AR = b²/S (S = area). Low AR is roughly b²/S < 2, moderate AR means b²/S around 5 and high AR is b²/S > 10. But there is no firm definition - every author is free to pick his or her own boundaries.

The center of pressure is defined as the point on the wing/airfoil where there is a zero net moment. This definition means that it is a function of the lift and subsequently of the lift coefficient. Airfoil theory overestimates the slope of the lift curve due to the absence of wing tip effects. With regards to the second question - I think the answer is rather subjective. Mathematically the limit of a very low aspect ratio is a 2D airfoil moving through unconstrained medium. How high an AR can go is mainly dictated by structural soundness of the wing. For real aircraft, have a look at the classic example of a low aspect ratio wing - F-104. Typical high aspect ratio aircraft are most gliders. I would point you to the exceptionally interesting read that is Theory of wing Sections by Abbott and Doenhoff and also any book on aerodynamics, e.g. Anderson


text:
Why is the coefficient of drag for straight wings at supersonic speeds lower than for swept wings?
Why is the coefficient of drag of a straight wing lower than the coefficient of drag of a swept back wing at higher supersonic speeds (above, say, Mach 2)?

Related, possible duplicate.

It's just logical. Swept back wing with the same wingspan has more surface area so more drag. Note that swept back wing of the same length as straight wing obviously has the same surface area but because it's swept back it will have less lift. So to increase lift we need to make the swept wing longer. When they're roughly the same span, generally they're also generating the same amount of lift. So the swept wing ends up having more surface area thus more drag.

@fooot, that question is about low speed, but this one is about high speed.

By "high mach numbers" you mean transsonic (say ~M0.7–M1.5), high supersonic (M2–M5) or hypersonic (>M5)?

At high supersonic speeds (M2-M5)

@SteveV.: Honestly, this is no duplicate. This is not about transsonic aerodynamics, but solid super- and near hypersonic speed. Now sweep becomes a liability, which goes counter the supposedly "original" answer. Please reopen!

After being edited, this question is no longer a duplicate.

@PeterKämpf, reopened.

@SteveV., it never was a duplicate. It asks why straight wings are better, which is not true at transsonic speed (and that's what the suggested duplicate explains). It is, however, true at higher supersonic. It was not very clear though.

Mark is right when he says that there is no induced drag at supersonic speeds, but it is an invitation for misunderstandings. Induced drag is replaced by lift wave drag, and all what happens is that aerodynamicists choose to use two different names for basically the same effect: Air gets pushed down.
As usual when I post a long answer, I was not quite happy with the existing answer(s). Now I have some time and try to give a better answer.
First, why does the straight wing work better only at very high Mach numbers (> 2.0)? Because at lower supersonic speed a swept wing gives overall better performance. The sweep angle $\varphi_0$ must be high enough to allow for a subsonic leading edge (Mach < $\frac{1}{cos\varphi_0}$). Then the flow around the leading edge is subsonic and creates a suction area when accelerating around the nose contour. This suction helps to reduce drag - after all, this same suction is why a subsonic airfoil in inviscid flow has no drag. Edward C. Polhamus did a lot of research on this at NACA Langley and published several papers with equations for calculating the suction force.
Once you fly faster than Mach 2, the sweep angle for a subsonic leading edge rapidly gets too high for acceptable subsonic flight, and an unswept wing becomes the better alternative since you need to accept a supersonic leading edge. Examples are the wing of the F-104 or the canard of the XB-70.
Now for the airfoil drag at supersonic speed. Because it is easiest to explain, I select a rhombic cross section:

Rhombic airfoil in supersonic flow at zero angle of attack (own work). The plus sign denotes higher pressure, the minus sign lower pressure than ambient. By selecting a rhombic airfoil, the flow is very easy to determine because the pressure only changes when the local contour gradient changes. The two compression shocks create the typical sonic boom when arriving at the ground. Note that this airfoil already creates pressure drag even at zero lift. Any airfoil thickness greater than zero and any airfoil camber will cause this type of drag where the forward-facing areas see higher pressure and the rear-facing areas experience suction. This type of drag is called wave drag. It can only be minimized by minimizing the relative thickness of whatever is supposed to fly at supersonic speed.
When the angle of attack is increased, this airfoil begins to create lift. Now the compression by the lower forward shock becomes stronger and that by the upper forward shock becomes weaker. The expansion fan is the same again on both sides, so the upper rear half experiences less pressure than the lower rear half. I tried to symbolize this by the amount of plus and minus signs:

Note that the pressure difference is constant over chord, so the center of pressure is at 50% of chord length. Note also that the lift vector is perpendicular to the chord line. Since lift is defined as the force perpendicular to the direction of the undisturbed air, supersonic lift always carries a drag component which is proportional to the angle of attack - there is no suction at the nose to alleviate this! The wave drag of the airfoil at zero angle of attack still comes on top, so we have a shape-dependent wave drag and a lift-dependent wave drag component. This lift-dependent wave drag replaces the induced drag of subsonic speeds. If we compare the magnitude of both, we find:
Subsonic: $c_{Di} = \frac{c^2_L}{\pi\cdot AR\cdot\epsilon}$
Supersonic formula for 2D flow: $c_{{DW}_L} = c_L\cdot\alpha$
This doesn't look so similar, so let's now express the angle of attack $\alpha$ by the lift coefficient divided by the lift curve slope:
$$\alpha = \frac{c_L}{c_{L\alpha}} = \frac{c_L}{\frac{4}{\sqrt{Ma^2-1}}\cdot\left(1 - \frac{\lambda}{2\cdot AR\cdot\sqrt{Ma^2-1}}\right)}$$
and the lift wave drag component becomes $c_{{DW}_L} = \frac{c^2_L}{\frac{4}{\sqrt{Ma^2-1}}\cdot\left(1 - \frac{\lambda}{2\cdot AR\cdot\sqrt{Ma^2-1}}\right)}$
Now let's compare the F-104 wing, which has an aspect ratio $AR$ of 2.45 and a taper ratio $\lambda$ of 0.385: If we plug in the parameters and adjust $\epsilon$ such that both sub- and supersonic lift-dependent drag coefficients agree, $\epsilon$ would need to be 0.89 at Mach 1.2, 0.58 at Mach 1.4 and 0.31 at Mach 2.0. The dramatic rise of lift wave drag over Mach is caused by the reduction in the lift curve slope over Mach.
For slender bodies the lift curve slope is $c_{L\alpha} = \frac{\pi\cdot AR}{2}$ and the lift wave drag component becomes $c_{{DW}_L} = 2\cdot\frac{c^2_L}{\pi\cdot AR}$. For slender bodies the supersonic $\epsilon$ is 0.5, regardless of Mach.
The important conclusions from this for wing selection are:

Sweep does no longer help once the leading edge is supersonic.
Lift wave drag continues where induced drag drops off. Lift always causes drag.
For a supersonic wing the aspect ratio is of minor importance.

Now back to the original question: Once the leading edge is supersonic, sweep is no longer helpful. Now the best wing is straight, because it will need the lowest wing area to create the required lift at subsonic speed. At supersonic speed its lower area will translate into lower friction drag, making it better than comparable delta or swept wings.

I really like the clarity of your illustrations. Do you have any links for further readings (and if not, I think you should publish a book compiling your answers)?

For a given lift coefficient, boundary layer drag is smaller for the straight wing configuration, and lift-induced drag is larger for straight wing.  These conflicting effects may give an advantage to straight wing at low subsonic speeds.
At speeds of about Mach 0.8 you begin to get wave drag because the flow over the wing is not uniform, and in some regions you have supersonic flow.  The wave drag is much less for swept wings, roughly in proportion to the aspect ratio.  So when you are trying to push the plane to Mach 1, swept wings make your job easier.
At Mach 1 and a bit higher more and more of the wing region experiences supersonic flow, and the shock drag is dominant.  Again, swept wings have a huge  drag advantage (not to mention a control advantage).
But pretty soon, certainly by Mach 2, both wings have the same amount of wave drag, and the shock drag becomes less important as the region near Mach 1 disappears.  And now we come to our first observation, about boundary layer and lift-induced drag roughly balancing.  Except --
At supersonic flow there is no lift-induced drag
This is because the "wake cannot be felt upstream."  More accurately, the penalty associated with turning the flow to generate lift is captured over the wing surface and what happens downstream of the wing cannot affect the flow over the wing because "knowledge" of what happens downstream propagates at the speed of sound.
So we have to remove, from our calculations for each type of wing, the lift-induced drag, which had been greater in the straight wing.  And this makes the straight wing drag coefficient lower at moderately high supersonic speeds, compared to the swept wing.
I may be wrong here because fluid dynamics is famous for introducing subtle effects that nobody would have anticipated, but I believe this issue is understood and the trade-off mentioned is the heuristic reason.


text:
Why does the MiG 21 use anti surge shutters?

Does the MiG-21 use anti surge shutters at high speed? What is the adverse effect on the airflow at such high speeds (despite using a conical spike ahead) which warrants the use of antisurge shutters?
Look just ahead of the wing root, below the two red triangles. The anti surge shutter is covered with a red coloured blanking secured with a yellow cord.

Please clarify what you mean: Is that the bleed valve ahead of the canopy? Maybe a picture or sketch would help - it is not clear to me for what detail on the MiG-21 the term can be applied correctly.

Peter Kampf, look just next to the ejection seat, below the two red triangles. The anti surge shutter is covered with a red coloured blanking secured with a yellow cord.

Mig 21 drawing should anybody find it useful. Part in question appears to be #45 labelled as a 'suction relief door'.

You should provide the source of your image.

Normally, such shutters are close to the intake and covered by simple spring-operated doors which open when the pressure in the intake duct drops below ambient pressure at low speed.

AV-8B air intake (picture source). The upper two doors are open while the lower three are closed.
The ideal area of a jet intake varies with flight speed. At low speed, it needs to be large so the amount of air swallowed by the engine can be supplied without too much resistance. At high speed, a big intake captures and compresses more air than the engine needs, which increases drag. Especially fighter jets need an intake which is optimized for high speed flight, and in consequence much of the engine power is needed to suck in enough air during the take-off run. Adding doors which are sucked open in that condition will add another pathway for air to reach the engine, which in turn will help to avoid a compressor surge and improves thrust at low speed.
I have no detail knowledge of the MiG-21 intake, but it appears to me that the anti-surge doors are only open at low speed and high engine RPM. At high speed, the pressure in the intake duct is a lot higher than the ambient pressure and the doors are firmly closed.

From:Mikoyan-Gurevich MiG-21 Pilot's Flight Operating Instructions
In order to operate at supersonic speeds, the engine inlet has a
movable nose cone that extends as the aircraft speeds up. This keeps
the shock wave in front of the inlet.
The engine of the MIG-21 is protected against compressor surges(a reversal of air pressures causes air to move in a wrong direction)by the anti-surge shutters that open to restore normal airflow in the engine. Exhaust is routed through a controllable nozzle that opens or closes to maximize engine thrust and smoothness.


text:
Are airline pilots worn out at an age of 55?
Lufthansa's pilot union is calling for a strike to defend pilot's right to retire at the age of 55. They argue that the stressful job of flying is too much for older people. This provokes two questions:

What fraction of pilots takes advantage of such a rule (if it is available to them)?
Is there really a steep decline in the capabilities of older pilots?

I would expect that many pilots continue to fly, so I wonder whether it makes sense for Lufthansa to risk full confrontation. Also, if the reasoning of the union is right, it would be prudent for airlines to retire older pilots before they are a safety risk.

it would be prudent for airlines to retire older pilots before they are a security risk Aging might affect different people in different ways? So one person might want to retire at 55, while another doesn't. (Captain Sullenberger was turning 58 in 2009).

There are many more reasons for this strike, pension age is only one of them.

Airline pilots don't all wear out at the same age. It depends on their number of flight cycles and the quality of preventative maintenance. Excessive lubrication leads to early retirement.

anecdotally, in the "for what its worth" department ... President Regan fired all FAA traffic controllers who went on strike. One of the controllers' union grievances was job stress. At the time a co-worker of mine - Joe - had a good friend - Mike - who was a FAA controller. Mike, prior to all the strike brouhaha, tells Joe that he should become an air traffic controller. Great job. Decent pay. Flexible hours. EASY. Time passes, and now the strike is imminent. One day Joe laughingly tells me how Mike is now whining about undue job stress, lagging pay, crummy hours, etc. etc.

I wonder why asking about retiring ages is considered opinion-based now. I would more be interested in hard facts, anyway.

Personally I abstained from voting, as I understood your intentions, but the wording felt borderline. You can try to explicitely ask for research data/official statistics, rather than leave the questions open-ended. But this could make answers such as Terry's "in my experience" out of scope.

I suggest splitting your two questions into two actually separate questions. Data regarding retirement ages of pilots is hard data, so that shouldn't be opinion based. Data on the ability of pilots to fly is a bit softer, but should not ncessarily be opinion based either. However, having those combined in one question may have lead some users to see this as an opinion based question, inviting speculation about pilot retirement.

@PeterKämpf: I felt it was a good question, hence voted for re-opening. If you could ask for some published statistics, as suggested by Federico, then I think this question has merit.

In which country? For which companies? For which generation of pilots? etc, results may be different and explained in the answers. VTR with that in mind.

My experience, speaking as a pilot who had to retire in 1999 from a small 747 carrier when the US age 60 rule was in effect, was that NONE of us close to retirement wanted to retire. More often than not, the attitude was, "It's hard to believe they actually pay us for this." We didn't want to quit; we were at the top of our game; the world was our playground. The money was also very good!
Insofar as worries of a "steep decline in capabilities of older pilots," there is no statistical evidence that I am aware of to support this, although whether there is at least some decline has been a matter of controversy. You can Google "age 60 rule" and find reams (screen loads?) of info on this. A particularly good link IMO is http://www.avweb.com/news/aeromed/181875-1.html
What makes a good air carrier captain is not so much being a good stick or quick reactions, but being situationally aware, learning what to avoid, being able to read the politics of a situation (especially in third world countries), and other things that come to one primarily through experience.
For example, when I hired on to my second 747 carrier as an f.o., I was paired with a relatively young, fresh captain. I was older, had been a 747 captain at a previous carrier, and had more hours in the airplane. We landed in Harare, Zimbabwe to pick up a charter. Our APU was inop, so as per standard procedure, we left #4 engine running to supply the pneumatic air we would need to start the other engines. However, when the ground handlers told us there would be a delay of a couple of hours, the captain ordered the engines shut down, saying that we would use a huffer to start. Both the f.e. and I simultaneously objected, both saying in so many words that we couldn't depend on there being a usable huffer in Harare. The captain ignored us and shut down the engines, believing that the capital of a country would surely have such. The fun started when he asked for a huffer to be hooked up for our start. Long story short, what would have been but a two hour delay turned into something like a 10 hour delay. It included hooking up 4 different beat up old huffers to the airplane, but even using the two best at once (the 747 can accommodate 2 hookups) failed. Finally, a mechanic worked on those two best units and managed to get enough pressure to start us.
The captain was a good stick; he just hadn't had enough experience.

Hi Terry, I was looking forward to your answer, and you didn't disappoint me! I asked to get the view of an expert like you, and fully agree with your view.

I'm disturbed that you think a plane that can't start its own engines is airworthy.

@Joshua Dispatching large aircraft with inoperative APUs is common. When the APU is not needed for flight (indeed in many cases cannot even be used in flight), airworthiness is not compromised. There are, in fact, many aircraft that have to have a ground-assist to start. Think of it this way, if performance once is the air is the overriding factor, why would you want to haul into the air all of the weight of the equipment that would allow you to self-start. The problem at Harare wasn't airworthiness, but rather "ground worthiness" if you will of the ground support equipment.

I think the core of the answer is here "The money was also very good!". Would you have felt the same if you would have been paid half or less? (that is the ~ situation today w.r.t. 1999)

@Federico: I'd think the money would be absolutely irrelevant to questions of proficiency, and fairly irrelevant to the retirement question.  I mean, if you were satisfied with making $X at age 59, why wouldn't you be satisfied making it at 61?  And if you weren't satisfied, wouldn't you have quit and gone into some more lucrative field long before you reached the age where retirement was an option?

@jamesqf because the people that now have 59 years and gain X, were very happy when years ago were getting 2X or 3X, and when the money got cut, they were too old to be employable elsewhere. I see this first hand and money is very much part of the question.

@Federico: I think you're addressing something else.  Of course people are seldom if ever happy with salary cuts, but I'm talking about pilots who were competent and satisfied to be flying for $X at age 59.  Why should that change just because they're a year or two older?  Or even if the option is to keep on flying for 1/2 X, rather than be forced into retirement.  Works that way for me, anyway (in a different line of work): I've several times taken lower-paying jobs because the work was more interesting.

@joshua, I have around 2800 hours in airplanes that needed a huffer to start.  Everyone of them was airworthy.  Well, most of the time...


text:
How are you supposed to get the East/West variation on this question in the FAA PPL written exam?
While taking a practice test, I needed to calculate the magnetic heading between two airports. Most of the time I run into a problem, the map will provide me with isogonic lines with the East/West variation visible. But in at least one or more questions, the isogonic line is either not shown, or the numbers are not visible.
Using the protractor to calculate the angle on the isogonic line using the latitude/longitude lines doesn't seem to give me the correct number either.
Does anyone know what I am doing wrong or how I can come to a correct answer?
Problem
(Refer to figure 23.) What is the estimated time en route for a flight from Claxton-Evans County Airport (area 2) to Hampton Varnville Airport (area 1)? The wind is from 290° at 18 knots and the true airspeed is 85 knots. Add 2 minutes for climb-out.
Valid Answers:

44
39
35

Update: I added an actual sample problem. Assuming the only information I am able to work with is the map, the question and my E6B (manual or electronic).
Update
I found the original question that got this thread started:
(Refer to figure 24.) Determine the magnetic heading for a flight from Allendale County Airport (area 1) to Claxton-Evans County Airport (area 2). The wing is from 090° at 16 knots, and the true airspeed is 90 knots.
The correct answer: 209°
The solution given in study mode: With a plotter, measure the true course from Allendale to Claxton-Evans as 212°. Use a flight computer to find the true heading of 203°. Add variation of 6°W. 203°+6°=209°.

VORs are aligned to magnetic north when they are commissioned and then periodically adjusted. You should be able to use the compass rose to determine offset. Also, I see the line on the east side of the map you posted, but don't see the printed offset.

Measuring angles on the map may be wrong because as the note says, the map is not to scale. It could be interesting you add more details about your test question (your location and the destination), so that practical answers could be proposed. From the longitude lines and SAV / ALD VORs 360° line the 5° degrees change is clearly visible.

By the way, I've seen this map "excerpt" before and there is an isogonic line just to the west of the Claxton airport with a 6°W variation marking.

@Ron Beyer I think, though, that the excerpt is all that is provided, so he has to use what's on the page

@Ron Beyer Also, when you say the VOR declination is "periodically adjusted" I think that period is probably pretty long since they would have to change all the airways, fixes, procedures, etc would have to be changed to compensate. I doubt they do that often

If you look at the V37 airway, it is on the 1° radial of SAV and the 176° radial of ALD, so the two VOR's disagree by 3°. And the isogonic line looks visually to be more than that. (It's 7°. I cheated and looked it up) You can't go by the VOR roses for an accurate magnetic heading.

Just a guess, as I don't know how you Americans do it - in Australia our airfield directory lists the magnetic variation at each strip. It would be a lot more convenient if they just gave you a complete excerpt but that approach should get you there.

I will try and find the original problem and add its details to the question. Assuming I can only use the material given to solve my problem during my test, the map doesn't appear to provide all the details necessary to accurately answer the question. I tried to use the VOR radials, but they weren't able to get me to the correct answer with any degree of accuracy.

@TomMcW the line is there, but the variation is not visible. The current map says 7° west. If I use my protractor to get the angle of the isogonic line, I get a number closer to 12°.

@Ben I am unable to use materials other than those they provide (and a flight computer) to take the test.

@David I can see the predicament. I was just pointing out that the solution of using the VOR's doesn't work. I'm watching this question to see if someone comes up with an answer because I'm wanting to know myself

The angle that the isogonic line makes to the meridians is not related to the variation.

The answer is that magnetic variation is not a factor in this problem.
See this answer:
When are winds given with respect to true vs. magnetic north?

Another potential non-real-world issue with this problem might come from this:

NOTE: Chart is not to scale and should not be used for navigation. Use associated scale.

So, use another piece of paper to figure out how far the two airports are, then use your plotter to find the True Heading-- do not use your plotter to measure the distance!

Here is my solution to your original problem. I'm not going to work out the second one since you already have the solution, and the a real question should include the variation as clearly readable information somewhere, either on the chart or in an AFD entry.
I am using a method that doesn't require writing on the chart, which is not allowed on the FAA tests. In real life, you would draw your course on the map for reference during your flight.
First, find the distance between the two points. Mark it on the edge of one of your blank sheets:

Now measure the distance using your on-maps scale not your plotter

I got 57 nautical miles, from the center of Claxton to the center of Hampton.
Estimate the true bearing from Claxton to Hampton. This is a northeast course, so we know the true course will be around 40-50 degrees.
Now put the Course Arrow on the two airports, and the hole of the plotter on the latitude line:

From your plotter, read off exactly 45 degrees from the third scale. You know to ignore the others because you expect the answer to be near 40 or 50 degrees.
Now pull out the whiz-wheel and follow the directions on it!

Set wind direction opposite TRUE INDEX
Mark W up from the Grommet (add 18kts)

Place true course under TRUE INDEX
Slide true airspeed under W (85kts)
Read Ground Speed under the Grommet

I get 91.5 kts GS.
Now you have all the information you need to do the calculation:
Remembering that a knot is a nautical mile per hour,
$${ 57nm \over 91.5kts } \times {60\;min \over hr} = 37.4 \; minutes$$
Now add two minutes for climb as directed in the problem, and you get a little over 39 minutes.
Note that you can ignore variation in this problem because both wind and course are calculated as true bearings. Since the problem doesn't ask what the Magnetic Heading or Course Heading are, that information isn't required.

I added more detail at the bottom of my question to show an example of where the map was used and variation was required.

@David It is highly unlikely that you'll encounter that in the exam, where basically a typo or figure update has invalidated a question.

thanks, it's good to know. It's hard enough to learn without confusing stuff like that making you think you are doing something wrong.

Thanks for the detailed answer. I think your comment that the problem is not likely to show up on the test makes the most sense.

Real answer: The FAA screwed up in preparation of that exam question (this sometimes happens; they aren't perfect) and cropped the sectional to where the isogonal lines and or isogonal values are not visible.  As it happens I know that area of the country pretty well.
It works out that Claxton has about a 6.5 deg W magnetic variation (get the full VFR sectionals for that area) and Varnville is just shy of 7deg  W variation; I'd use 6.75 deg W as your number for mag variation over the duration of the flight.  That works out to a course of 051 deg magnetic, 56.5 NM along the route.  With winds at 290 at 18 kts cruising at a true airspeed of 85 kts, this gives an aircraft heading of 041 magnetic in cruise with a ground speed of 93 kts.
Now here's the thing: what is your groundspeed during climb out?  That's going to also be necessary to calculate how much ground you cover during your 2 min climb to cruise. And critical to answering the question correctly.
Now, IF WE ASSUME a constant groundspeed of 93 kts, from point to point, that gives a flight time of 36.5 minutes.  If we further assume the two minutes of climb are not enroute and just maneuvering from takeoff to on course right over the Claxton airport and we descent into Varnville at 93 kts as well, we get a figure of 38.5 minutes total flight time.  Therefore, I'd select 'B' as my answer.
As a side note, I landed an LSA out at Varnville only once and that place is virtually abandoned; it was difficult to identify as an airport from the air, the runway is anemic and derelict, there are virtually no buildings there and all I could find for traffic there was a decrepit C-150 rusting in the sun.  I took off again - and very glad my engine didn't quit, leaving me stranded there.

I think this answer is misleading because during the pre-flight phase, you would be acquiring your wind information from a briefing weather source, which is given in true rather than magnetic. The question doesn't ask for the Magnetic (course) Heading, so variation is not needed. Since the variation is so small near Savannah and the trip is so short though, it doesn't significantly affect the answer even if you incorporate it.

You should differentiate between "critical to answering this question correctly" and knowing the answer for your actual XC flight (ie not the written exam). These kinds of questions seem to only involve the distance between two points, a simple, constant wind, a simple, constant speed, and a small 'rule of thumb' adder (in this case, 2 minutes). The ground speed during climb is factored in for you via the 'rule of thumb' number -- it's not any more complicated than that.

The reason for trying to get to the bottom of this answer is I want to do well on the written exam. I know the questions isn't exactly a real scenario, but in order to pass the test, you have to get the answers correct. Several questions in the Sporty's practice exam use that map. The solutions they provide show you including the variation in the calculation, but the map does not provide that information. I was wondering if they messed up or I was blind. As a new pilot, I couldn't say for certain. Thanks for all the feedback.

These are surprisingly close to a real scenario though! My point about the rule of thumb is that you'd need take into account your aircraft climb performance, etc in the real world, but on the test they make it easy with the 'add 2 minutes for climb' clause. A possible reason for this figure weirdness is that this question references 'Figure 24'... that's obviously not right, so perhaps it refers to an old edition of the Airman Testing Supplement... On the real test it seems unlikely that you'll run into this bad of a whopper!

That's my point, David.  The guy who wrote that test question screwed up and cropped the image such that you can't see the isogonal lines or numbers.


text:
What is the effect of changing the autobrake setting after touchdown?
If an autobrake setting is chosen prior to touchdown, but after touchdown and nosewheel compression (i.e. after the autobrake has engaged) a different setting is required, is it safe to change the setting during the roll-out? What happens when the setting is changed in this manner? I know application of manual brakes disables the autobrake. Does the autobrake immediately match the new deceleration requested?

I researched this a bit and couldn't find a definitive answer. The Boeing 737 flight manual (only aircraft with autobrakes whose manual I had easy access to) does not specify what happens; it just says that autobraking will cease if the knob is moved to the OFF position. I can tell you that at least one 737 simulator will adjust braking force if you change the setting during rollout, but I can't find anything confirming that the aircraft does this. (Sims aren't always perfect!)

Assuming it has been actually deployed (I could not find any articles or documents newer than 2010, so could not confirm it's actually in use) the Airbus A380 has the (new) BTV or Brake-To-Vacate system. The pilots code in where they will be planning to exit the runway and the system will automatically adjust braking to ensure the plane can turn at the planned exit. If it cannot (with limits of temp, etc) the system is designed to audibly warn the pilots that the "runway is too short" to stop in the distance allotted. This doesn't answer the OP per se, so I've left it a comment.

Thought I would contribute something more difficult as I expect others will contribute to some of my difficult questions as well. I am not a pilot and know very little about flight operations but am in a position to dig up this sort of stuff. Hope it makes sense from a pilot's point of view.

This answer was difficult to find as both Flight Operation Manuals and Maintenance Manuals did not say. The only guidance was found in the system description provided by the OEM of the BCSU.  I am not mentioning the fleet type because at some point the detailed description might become proprietary.
According to the description I found for a given automatic braking system is that it is armed when the pilot presses one of the three pushbutton switches which set the auto-brake level.
The system will only arm if gear up is not commanded, there is hydraulic pressure as needed, and no existing faults exist that prohibit normal braking, as well as a few other computers available on the associated data bus.
There is nothing that says the system can't be armed after touchdown, however the system will only engage when the spoilers are being activated, or are activated. Or more precisely from the black-box point of view automatic braking is engaged whenever two spoiler signals in three appear once automatic braking has been armed.  In other words, once armed, the automatic braking will only engage if spoilers are sending the signal that they are being activated, or are activated.
It specifically says that the pilot can select another pushbutton switch even after
arming. The BSCU acknowledges this new command and executes
the corresponding task.
Automatic braking is released whenever the braking sequence
is interrupted:

when an arming condition disappears,
when an engaging condition disappears,
when one or both brake pedals are held down beyond the
defined threshold,
when the selected pushbutton is pressed again.

The question, therefore is 'After you have already armed at a given setting and the system is actually engaged, because the spoiler actuation has initiated auto-braking, what happens when you press another setting?'  As long as the arming conditions still exist, which in most cases would, and as long as the engage conditions exist, as would be the case if the spoilers are activated (at least 2 of three spoiler signals being sent to the BSCU), then yes, you may change the setting after touchdown and the BSCU acknowledges this new command and executes the corresponding task.

This would work for an Airbus in a discrete level of braking (i.e. 1, 2, or 3.) but not in BTV (brake-to-vacate) since in that mode the computers are constantly changing the braking levels as needed to slow the plan to a speed at which it can safely exit the runway, given the angle of exit, and that the brakes remain under the threashold temperature.

@CGCampbell - Agreed - with new models that have BTV there are other issues to consider.

This will be quoted from the B737 NG - Systems Summary [Landing Gear] which isn't an official Boeing document, but ...
Page 5

Landing
  When a landing autobrake selection is made, the system performs a turn-on-self-test. If the turn-on-self-test is not successful, the AUTO BRAKE DISARM light illuminates and the autobrake system does not arm.
Four levels of deceleration can be selected for landing. However, on dry runways, the maximum autobrake deceleration rate in the landing mode is less than that produced by full pedal braking.
After landing, the autobrake application begins when:
   - both forward thrust levers are retarded to IDLE
   - the main wheels spin-up
NOTE: Landing autobrake settings may be selected after touchdown prior to decelerating through 60 kts of ground speed. Braking initiates immediately if the above conditions are met.
To maintain the selected landing deceleration rate, autobrake pressure is reduced as other controls, such as thrust reversers and spoilers (without disarming the system) by rotating the selector. The autobrake system brings the airplane to a complete stop unless the braking is terminated by the pilot.

So, reading between the lines here, if the autobrake system has not been activated, upon touchdown it can be activated, before slow down to 60kts ground speed, as long as all other stipulations are met. Also, although the sentence as written, is confusing, the last paragraph indicates that the level of braking can be changed during autobraking, by turning the selector AND that will not disarm the system.
I know this is not official documentation and is an older model plane, but thought I'd offer it up.

I have flown the 737NG and now fly 757-200. It is now quite normal for us to set a certain AB setting (e.g. level 3), to match a desired runway turn off.  Then once landing distance is assured, at maybe 60-100kt, reduce the setting to 2, then 1, then off (or disarm on 757).
Alternatively you can cancel AB by using manual braking pressure (normal method on 737) or by moving the speed brake lever slightly forward, then back to fully deployed once AB have clicked out (normal method on 757).
All methods work on both aircraft, and the choice of method is more about the best way of avoiding the slight jolt felt as the ABs disengage - it's all about passenger (or pilot!) comfort - oh yes, and also about stopping by the end of the runway!

I generally use AB 3 on the B767 and B777 and will sometimes reduce to 2 or 1 during the roll out.

Yes the autobrakes can be increased after landing using the selector switch but it depends on pilot preference and certain conditions.

You may precise your answer by preceising on which aircraft type this maneuvre works, by precisung the words "certain conditions" and "pilot  preference", and by adding references and links for further reading


text:
What are Electro Hydraulic Servo Valves and Solenoid Operated Valves?
What is Electro Hydraulic Servo Valve (EHSV) and Solenoid Operated Valve (SOV)? Also what is the difference between these two valves?
Basically, these two are used to control the hydraulic fluid in actuators. I am working on a aircraft program where SOV are used in Aileron, Elevator & Rudder actuators while EHSV are used in Multi-Function Spoilers [or just spoiler]. I think someone explains these two valve basics would provide me the reason for the selection of these valve to control the hydraulic actuators.

Someone else can probably provide a better answer, but the short version is that a SOV is either open or closed, providing on/off control for something while the EHSV can be commanded to be fully open, fully closed, or anywhere in between (usually via a variable voltage input or "pulses" of electricity to tell it to open or close by a fixed amount), providing control over "how much" instead of just on/off.

A solenoid is an induction-based actuator. In its simplest form it is a magnetic rod inside a coil of wire. Run a current through the wire and you create a magnetic field that creates a force on the rod, moving it one direction or the other. The force with which this occurs is proportional to the current through the inductor (audio speakers use this exact same principle to make the cone vibrate).

A solenoid-operated valve usually functions as a simple gate; the solenoid mechanism pulls or pushes a simple flat valve allowing the fluid to pass through or not. It's technically possible, but not usually feasible, to use a solenoid as a variable valve; for it to be accurate would require very low friction so the current applied to the solenoid (and thus the force on the valve) could be varied to produce smooth variations in valve position.
Solenoid-operated valves are typically used as triggers for larger mechanisms, including the electrohydraulic servo valve.
An electrohydraulic servovalve uses electrical current to move a "flapper", similar to a solenoid. The flapper controls the flow of hydraulic fluid to one side or the other of a spool valve. That moves the spool valve, allowing the supply pressure to feed an actuator cylinder, moving it one way or the other. However, that same movement also allows hydraulic pressure to return up to the flapper, providing a countering hydraulic force to the electrical force that moved it, called "negative feedback". Eventually these forces balance, and the flapper re-centers, closing the spool valve. The net result is that the actuator cylinder has moved a certain set distance, proportional to the strength of the input current.

These valves are used to provide smooth, high-strength motion in a hydraulic system. This type of valve, for instance, can be used to move the control surfaces of a large airliner, without requiring high-pressure hydraulic lines or tensioned cables to link the hydraulic system to the yoke in the cockpit. Instead, the yoke (or sidestick) controls the current through electrical wires leading to the EHSVs. That link is much easier to maintain (and cheaper to build in the first place) than older systems on previous generations of aircraft.


text:
What areas of a commercial airliner are pressurized besides the cabin?
For example, what about the baggage compartment? Landing gear housing? Avionics?

Question for cargo compartments.

@fooot, my question is more general, i.e. not asking only about cargo compartments.

Cockpit, yes!!!

Baggage, yes, landing gear, no.
The fuselage of an airliner is typically "capped" by two bulkheads, the forward and aft pressure bulkheads. These are roughly spherical sections that seal the central "cylinder" of the aircraft's fuselage.
Here's a 787's forward bulkhead:

... and a 767's aft pressure bulkhead:

Practically all the space in the fuselage itself between these two bulkheads is pressurized during high-altitude flight. This includes the baggage compartment.
Gear assemblies, however, are typically not pressurized. For some jets like the 737, this is fairly obvious:

Other designs with full doors for the gear assemblies aren't as obvious, but in general it's neither feasible nor desirable to pressurize the landing gear housings when airborne.

So basically everything inside the fuselage and between the two caps?

Save for the L/G bays, yes.  (This means that the holds are pressurized, and so are the avionics bays)

Not all the avionics, there is usually a radar array in the nose to provide TCAS and TAWS.

@KeithS, Neither TCAS nor TAWS uses a radar. TCAS uses an antenna since it is just a radio. TAWS is simply a database that is loaded into a display (often an FMS). The radar in the nose is simply for weather.

@KeithS, can you incorporate the info in these comments in your answer?

All the fuselage volume is at the same pressure because there is no separation between areas, for instance between the passenger cabin and the cargo holds. Circulation is facilitated by design and by regulation to protect the fuselage and prevent the cabin floor to collapse, particularly in case of explosive decompression.


text:
Does a Chicago to UAE flight fly over the Arctic Circle or go around it?
I'm speaking about a flight from Chicago to UAE. Does it fly over the Arctic Circle or go around it?

Are you asking about specific flights, or flights at high latitudes in general? If in general; why should there be anything special about the arctic circle? There's plenty of airports north of it.

Am speaking about a flight from Chicago to UAE

Have you had a look at the maps?

A direct flight from Chicago to the UAE will come close, but not enter the Arctic Circle.
Some flights will take a slightly more northerly route due to winds and weather conditions, but most will not enter the Arctic Circle.


text:
Why would a flight from North America to Asia sometimes fly over the Atlantic?
I am going to take AC031 from Toronto to Beijing in a couple of weeks, I searched its flight path, most of the time it will fly westbound after taking off from Toronto and continue west across the north pole area and land in Beijing, but yesterday it flew eastbound, across the Atlantic, this is very strange since the distance is a little bit more than flying westbound.
Why?

Very weird definition of "the Atlantic" you have there. It clearly flies over the north pole, not the Atlantic.

Comments are not for extended discussion; this conversation has been moved to chat.

It's worth noting that only the green segments on the flight path are from actual radar tracking; the white segments are just estimated paths, and are (I believe) just great-circle routes that connect the known positions.  So the actual routes for both flights might have been somewhat different.

I don't think it's really going east or west much, either way.  It flies north, then south :-)

Very weird definition of "clearly" you have there.  The problem is the map. Mercator projections are exceptionally poor for anything within a few thousand miles of a pole.

To complement the posted correct answers, consider this page: https://en.wikipedia.org/wiki/Antipodes#Cities It contains a list of cities that are exactly (or almost exactly) opposite one another on the spherical earth (antipodal); for example, Xi'an, China and Santiago, Chile. You could leave Santiago, and fly in a straight line (great circle, not a rhumb line) in literally any starting direction and still wind up close to Xi'an. Of course, range and availability of alternate landing fields along the way would limit your choices...

Spheres are weird.

Never use a flat map when thinking about a globe. The routes are straight, it's the map what's curved.

@jwenting over, as in North of, is what I think OP was saying. Haha

@Harper: Yes, they are indeed. But the very fact that we are talking about regions very close to the pole makes it clear that it is not the Atlantic.

@Yifan Nie take a look at Gnomonic projection of the world. you can understand the flight paths easily and its accurate.

Get a globe and stand over it and look adown at these two routes drawn on the globe. Then you will understand.

Those routes are extremely similar.
They're basically two almost-a-straight-line routes over the arctic ocean, except that the second one has deviated a bit to the right, so that it's gone just on the right of the North Pole instead of passing by the left. At the pole itself, there's no such thing as "East" or "West", and if you pass very close to it, there is a discontinuous jump between passing it "going East" and "going West".
You can see this explicitly by going to FlightAware, asking to View track log, and downloading the flight tracking data to be viewed on Google Earth:

 Actual tracking data in white. Interpolations for the missing-data parts in green.
As you can see, there is indeed a nontrivial deviation of what's probably several hundred kilometers in clipping Nunavut vs the northeast corner of Greenland, but that deviation is within the normal ranges of what you'd get in your standard transatlantic flight, say. The only change here is that the variation now includes the North Pole itself.

Credit obviously due to the three answers posted previously to this one, which contain very similar content - I just thought I could provide a slightly clearer version and pull in the actual flight tracks.

Unrelated to the OP's question: is it possible one or the other of those (similar) routes is chosen due to eg. airstream etc ??

@Fattie There are many factors that go into planning flight paths; wind conditions are obviously one of the most important ones.

It makes sense when you put it that way @E.P.  :)

Is there any way to overlay the two routes onto the same image?

@jpmc26 I tried, and failed, to get Google Earth to do that.

I took Hewgill's picture and added the routes in the OP (I just eyeballed this, so I won't guarantee accuracy).

With the FlightAware maps it looks like drastically different routes, but from this angle you can see that both routes are not that far off the ideal route.

This is actually the most useful answer as it shows visually how little difference there is between the flight tracks. Not that any of the others are wrong...

The direct route from YYZ to PEK flies almost over the north pole:

Sometimes, routes need to deviate a bit from the "direct" route, due to prevailing winds or other traffic. If your route from YYZ needs to deviate a bit to the right, then it will cross over to the other side of the north pole (which is the very centre of the map above), and your projected map from Flightaware would show it flying "over" the Atlantic. But really, it's just gone slightly to the right of the pole.

Showing the right picture can make a big difference. +1.

The aircraft must remain within 180 mins of a suitable landing airports at all times. There are very few airports along that Polar route so weather conditions may dictate a more easterly route over Greenland on some days.

The difference between 'east' and 'west' routes is almost always driven by the winds. The 'east' route is more common in the winter due to the stronger jet stream.

180 minutes - wouldn't that be about 1500 miles?  With airports in Alaska, Iceland, and Norway that doesn't look like it would be too difficult.

Not sure which is the relevant regulatory body here, but the Boeing 777-300ER has been certified by FAA in the US to fly as far as 330 minutes from airports http://aviationweek.com/awin/faa-extends-777-etops-approval

IDK what the A330's certification is, but the similar route DTW-PEK/PEK-DTW used to be flown by an A332 by Delta (Now they have A350 on it) and it followed approx the same route, basically over the pole.

@nasch No--that 180 minutes is with an engine out.  Look up the ETOPS rules.  (Which doesn't actually mean Engines Turn Or Passengers Swim.)

The flight did not fly across the Atlantic. Rather, you're seeing the results of stereographic projection.
More specifically, the Earth is Spherical. Navigation needs to be thought of on the basis of a sphere. It is 3-dimensional, and if you have studied sphere geometry, you will know that it defies the rules of geometry we know in 2D - for example it is possible to construct a triangle with three 90-degree angles.
The problem is that 3-dimensional objects cannot be displayed on a map, which is 2-dimensional. Therefore a mathematical formula is used to make a sphere look like a rectangle. The conversion is really skewed for regions near the poles. As a result, you cannot compare distances by measuring them on the projected map.
The route merely deviated slightly to the left in the second picture. However it is still flying over the North Pole, not the Atlantic. The deviation may be due to changes in winds aloft and/or traffic.

How is such a big curve (in the OP) only a 'slight' deviation? I don't understand how that works. It looks like they went far, far from the ideal path. ..

@Cloud That's a byproduct of the map projection used by FlightAware. The Earth isn't flat, so the shortest distance between two points won't show up as a straight line on that map (the actual shortest distance for flight planning purposes also takes winds into account). The ideal path isn't anything remotely close to a straight line.

@Cloud It only looks like a big curve due to the limitations of drawing that type of projection. It isn't nearly as bad as it looks.

@Cloud This great circle visualization of the ideal route might help you understand. Look at the 3D view to see how the route looks on a globe, then switch to the 2D view with the button in the top right and see how different it looks on a flat map.

@kwc Thanks, that did help

Some years ago, talking with a friend, I did a demonstration with a small ball and a tape measure.  (He worked in a custom tailor shop in Bangkok.)  The look on his face as he realized that the curved path really meant was PRICELESS.

Wait, wait, wait! You mean the earth is ROUND?! So, why would NBA players mislead us?

@TomMcW it's just a conspiracy theory - follow your NBA stars for true enlightenment, (but don't let ever let them navigate a ship/plane:).

Neither of maps here are in stereographic projection. The maps in the OP's post are in Mercator projection; the one from gcmap.com in Greg Hewgill's map is polar orthographic (you can see how the shapes of India and the Arabian Peninsula are distorted); and what E.P. uses looks like a perspective or fisheye projection seen from about one earth radius above the North Pole..

ALL flight paths on the earth are curved, cause the earth is a sphere! Which way that curve appears depends on which side of the curve you are observing it from. If you observe it frm south of the flight path it will appear to curve northwards and then south. If you observe it from East of the flight path it will appear to curve West and then East, etc.